I'm trying to solve this problem: Let $\alpha>0$ and $a_{n}=\frac{\cos n}{2n^{\alpha}}$ for all $n\in\mathbb{N}$. Prove that the series $\sum_{n=1}^{\infty}a_{n}$ coneverges.

For $\alpha>1$, we have that $\mid\cos x\mid\leq1$ for all $x\in\mathbb{R}$ and therefore for all $n\in\mathbb{N}$: $$ \left\lvert \frac{\cos n}{2n^{\alpha}}\right\rvert=\frac{\lvert\cos n\rvert}{2n^{\alpha}}\leq\frac{1}{2}\cdot\frac{1}{n^{\alpha}} $$

Using the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}$ is convergent for $\alpha>1$, we have for the comparison test that $\sum_{n=1}^{\infty}a_{n}$ is absolutely convergent for $\alpha>1$. And with this $\sum_{n=1}^{\infty}a_{n}$ is convergent. But I don't know how can I prove that $\sum_{n=1}^{\infty}a_{n}$ converges when $0<\alpha<1$. Could you help me or give me some idea to prove this?

  • $\begingroup$ Are you sure it converges? $\endgroup$ – herb steinberg Aug 19 '19 at 19:58
  • $\begingroup$ Use an Abel summation, knowing that the partial sums of $\cos{n}$ are bounded. $\endgroup$ – Mindlack Aug 19 '19 at 19:59

In the following, we derive Abel's test and Dirichlet's test for testing the convergence of infinite series in the form $\sum_{n=1}^{\infty}a_{n}b_{n}$. Firstly, we introduce Abel's transform, which is the discrete version of integration-by-part. Then we apply it to the difference of partial sums $S_{n+k}-S_{n}$.

Section 1: Abel's transform. Let $(a_{n})$ and $(b_{n})$ be sequences of real numbers. For each $n\in\mathbb{N}$, let $B_{n}=\sum_{k=1}^{n}b_{k}$. Define $B_{0}=0$. Let $N\in\mathbb{N}$. We rewrite the partial sum $\sum_{n=1}^{N}a_{n}b_{n}$ as \begin{eqnarray*} \sum_{n=1}^{N}a_{n}b_{n} & = & \sum_{n=1}^{N}a_{n}(B_{n}-B_{n-1})\\ & = & \sum_{n=1}^{N}a_{n}B_{n}-\sum_{n=1}^{N}a_{n}B_{n-1}\\ & = & \sum_{n=1}^{N}a_{n}B_{n}-\sum_{n=1}^{N-1}a_{n+1}B_{n}\\ & = & a_{N}B_{N}-\sum_{n=1}^{N-1}(a_{n+1}-a_{n})B_{n}. \end{eqnarray*} This is known as Abel's transform. Comparing to the well-known integration-by-part formula $\int_{a}^{b}f(x)g'(x)dx=[f(x)g(x)]_{a}^{b}-\int_{a}^{b}g(x)f'(x)dx,$ we note that $(a_{n})$ is analog to $f$ while $(b_{n})$ is analog to $g'$.

Section 2: Define partial sum $S_{n}=\sum_{k=1}^{n}a_{k}b_{k}$. Let $n,k\in\mathbb{N}$. We consider $S_{n+k}-S_{n}$. Applying Abel's transform, we have \begin{eqnarray*} S_{n+k}-S_{n} & = & \sum_{i=n+1}^{n+k}a_{i}b_{i}\\ & = & a_{n+k}(B_{n+k}-B_{n})-\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})(B_{i}-B_{n}). \end{eqnarray*} Now, we are ready to state and prove the following two theorems.

Theorem 1: If $(a_{n})$ is monotone and bounded, and the series $\sum_{n=1}^{\infty}b_{n}$ is convergent, then the series $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent.

Proof: Choose $M>0$ such that $|a_{n}|\leq M$ for all $n$. Let $\varepsilon>0$ be arbitrary. Choose $N$ such that $|B_{n+k}-B_{n}|<\varepsilon$ whenever $n\geq N$ and $k\in\mathbb{N}$. For any $n\geq N$ and $k\in\mathbb{N}$, we have that: \begin{eqnarray*} |S_{n+k}-S_{n}| & \leq & |a_{n+k}(B_{n+k}-B_{n})|+\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}||B_{i}-B_{n}|\\ & \leq & M\varepsilon+\sum_{i=n+1}^{n+k-1}\varepsilon|a_{i+1}-a_{i}|\\ & \leq & M\varepsilon+\varepsilon|\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})|\\ & \leq & M\varepsilon+\varepsilon|a_{n+k}-a_{n+1}|\\ & \leq & 3M\varepsilon \end{eqnarray*} This shows that $(S_{n})$ is a Cauchy sequence and hence $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent. In the above, we have used the fact that $(a_{n})$ is monotone to obtain $\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}|=|\sum_{i=n+1}^{n+k-1}(a_{i+1}-a_{i})|$.

Theorem 2: If $(a_{n})$ is monotone and $a_{n}\rightarrow0$, and $(B_{n})$ is bounded (where $B_{n}:=\sum_{k=1}^{n}b_{k}$), then the series $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent.

Proof: Let $M>0$ be such that $|B_{n}|\leq M$ for all $n$. Let $\varepsilon>0$ be arbitrary. Choose $N$ such that $|a_{n}|<\varepsilon$ whenever $n\geq N$. For any $n\geq N$ and $k\in\mathbb{N}$, we have: \begin{eqnarray*} |S_{n+k}-S_{n}| & \leq & |a_{n+k}(B_{n+k}-B_{n})|+\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}||B_{i}-B_{n}|\\ & \leq & 2M\varepsilon+2M\sum_{i=n+1}^{n+k-1}|a_{i+1}-a_{i}|\\ & = & 2M\varepsilon+2M|a_{n+k}-a_{n+1}|\\ & \leq & 6M\varepsilon. \end{eqnarray*} Therefore $(S_{n})$ is a Cauchy sequence and hence the series $\sum_{n=1}^{\infty}a_{n}b_{n}$ is convergent.


Now, we go to prove that for each $\alpha\in(0,1)$, the series $\sum_{n=1}^{\infty}\frac{\cos n}{2n^{\alpha}}$ is convergent.

Proof: Let $a_{n}=\frac{1}{2n^{\alpha}}$, $b_{n}=\cos n$. Clearly $(a_{n})$ is monotonic decreasing and $a_{n}\rightarrow0$. Let $B_{n}=\sum_{k=1}^{n}b_{k}$. We go to show that $(B_{n})$ is bounded, then apply Theorem 2...

We go to compute $\sum_{k=0}^{n-1}\cos k\theta$ and demonstrate the use of complex numbers. Let $\theta\in\mathbb{R}$ be such that $\cos\theta\neq1$. Define $z=\cos\theta+i\sin\theta$. By DeMoivre Theorem, $z^{k}=\cos(k\theta)+i\sin(k\theta)$ for any $k\in\mathbb{N}$. On one hand, \begin{eqnarray*} 1+z+z^{2}+\ldots+z^{n-1} & = & \frac{1-z^{n}}{1-z}\\ & = & \frac{1-(\cos n\theta+i\sin n\theta)}{1-(\cos\theta+i\sin\theta)}\\ & = & \frac{2\sin^{2}(\frac{n\theta}{2})-2i\sin(\frac{n\theta}{2})\cos(\frac{n\theta}{2})}{2\sin^{2}(\frac{\theta}{2})-2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}\\ & = & \frac{\sin(\frac{n\theta}{2})\left[\sin(\frac{n\theta}{2})-i\cos(\frac{n\theta}{2})\right]}{\sin(\frac{\theta}{2})\left[\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})\right]}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\frac{\left[\sin(\frac{n\theta}{2})-i\cos(\frac{n\theta}{2})\right](i)}{\left[\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})\right](i)}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\frac{\cos(\frac{n\theta}{2})+i\sin(\frac{n\theta}{2})}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\ & = & \frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cdot\left[\cos(\frac{n\theta}{2}-\frac{\theta}{2})+i\sin(\frac{n\theta}{2}-\frac{\theta}{2})\right]. \end{eqnarray*} On the other hand, \begin{eqnarray*} 1+z+z^{2}+\ldots+z^{n-1} & = & \sum_{k=0}^{n-1}\cos k\theta+i\sum_{k=1}^{n-1}\sin k\theta. \end{eqnarray*} Comparing the real part, we obtain: $$ \sum_{k=0}^{n-1}\cos k\theta=\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\cos\left(\frac{(n-1)\theta}{2}\right). $$ If $\cos\theta=1$, then $\theta=2m\pi$ for some $m\in\mathbb{Z}$. It follows that $\cos k\theta=1$ for all $k\in\mathbb{N}$. Therefore $\sum_{k=0}^{n-1}\cos k\theta=n.$ We conclude that: If $\theta\in\mathbb{R}$ such that $\cos\theta\neq1$, then $$ |\sum_{k=0}^{n-1}\cos k\theta|\leq\left|\frac{1}{\sin(\frac{\theta}{2})}\right|. $$ Hence, the partial sum of the infinite series $1+\cos\theta+\cos2\theta+\ldots$ is bounded by $\left|\frac{1}{\sin(\frac{\theta}{2})}\right|$. Clearly $\cos1\neq1$, so $(B_{n})$ is bounded.



Dirichlet’s test may be your friend.


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